Use the differential equation and the given point to find an equation of the function. The point is (-1, 3) dy/dx=-8/(2x+3)^3

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`dy=(-8)/((2x+3)^3)dx`  Integrate both sides:

`y=int (-8)/((2x+3)^3)dx`

Let `u=2x+3` ; then `du=2dx` Then we have:

`y=-4 int (du)/u^3=-4 int u^(-3)du`

`y=-4[-1/2 u^(-2)+C_1]` ** `C_1` a constant

`y=2u^(-2)+C`  ** `C=-4C_1` ** Substitute for u:

`y=2(2x+3)^(-2)+C` or `y=2/((2x+3)^2)+C`

Now we have the point (-1,3), so we can solve for the constant:




Thus the solution is `y=2/((2x+3)^2)+1`


To check, we can differentiate:


     `=-4(2x+3)^(-3)(2)`   Using the product and chain rules


     `=(-8)/(2x+3)^(3)` as required.

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